The angle bisector theorem and the midpoint theorem

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• (a) State and prove the angle bisector theorem: the bisector of an interior angle divides the opposite side into parts proportional to the other two sides.
• (b) Prove that joining the midpoints of two sides of a triangle gives a segment parallel to the third side and congruent to half of it.

Solution

(a) Angle bisector theorem. In triangle $ABC$ the bisector of $\hat A$ meets $BC$ at $D$; claim: $\dfrac{BD}{DC}=\dfrac{AB}{AC}$. Proof. Draw through $C$ the parallel to the bisector $AD$, meeting line $AB$ at $E$. Since $AD\parallel CE$:

• $\widehat{BAD}\cong\widehat{AEC}$ (corresponding) and $\widehat{DAC}\cong\widehat{ACE}$ (alternate interior).

But $\widehat{BAD}\cong\widehat{DAC}$ (bisector), so $\widehat{AEC}\cong\widehat{ACE}$: triangle $ACE$ is isosceles with $AE\cong AC$. Applying Thales’ theorem to $AD\parallel CE$ cut by $BE$ and $BC$: $\frac{BD}{DC}=\frac{BA}{AE}=\frac{AB}{AC}.\qquad\blacksquare$

(b) Midpoint theorem. Let $M,N$ be the midpoints of $AB$ and $AC$; claim: $MN\parallel BC$ and $MN=\dfrac12 BC$. Proof. Triangles $AMN$ and $ABC$ share angle $\hat A$ with the including sides in proportion: $\frac{AM}{AB}=\frac{AN}{AC}=\frac12.$ By the SAS similarity criterion $AMN\sim ABC$ with ratio $\dfrac12$. Hence $\widehat{AMN}\cong\widehat{ABC}$ (so $MN\parallel BC$) and $MN=\frac12\,BC.\qquad\blacksquare$ $\boxed{\dfrac{BD}{DC}=\dfrac{AB}{AC};\qquad MN\parallel BC,\ MN=\tfrac12 BC.}$